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First consider unit requirements, $A_{jr}\in \{0,1\}$. The Erlang loss formula for a resource with capacity $C$ and offered traffic $a$ isThe Erlang fixed point approximation assumes that resources block independently and that the traffic retained after screening at other resources can be treated as a Poisson process. If $B_j$ is the approximate blocking probability and $q_j=1-B_j$, the reduced-load approximation givesThe link whose load is being calculated is excluded from the screening product. Otherwise one would confuse its offered load with its carried load.

For existence, these equations define a continuous map from $[0,1{]}^J$ into itself. The Brouwer fixed-point theorem gives a fixed point. Assume $C_j\ge 1$; a zero-capacity resource forces rejection of every call needing it and can be removed together with those call types.

For uniqueness, let $K$ have probabilities proportional to $a^k/k!$, $0\le k\le C$, and write $m_C(a)=\mathbb{E}K$. Direct differentiation givesThese identities hold for $a>0$, and $m_C(0)=0$. Also $E(a,C)$ increases from zero to one and $m_C(a)$ increases from zero to $C$.

Put $p_j=-\log q_j$. Define $a_C(p)$ by $E(a_C(p),C)=1-e^{-p}$, and define $h_C(p)=m_C(a_C(p))$. This is a continuous, strictly increasing function on $[0,\infty )$, starting at zero and tending to $C$. Multiplying each Erlang fixed point approximation load equation by $q_j$ transforms it intoThese are exactly the zero-gradient conditions ofEach integral is a strictly convex function; the exponential terms are convex functions. Thus $\Phi$ is a strictly convex function. It is also a coercive function, since $h_{C_j}(p)\to C_j>0$, so it has a unique minimizer. At $p_j=0$, its partial derivative is negative if some positive-traffic route uses $j$, so that coordinate of the minimizer is positive. If no route uses $j$, the unique minimizing coordinate is zero. The minimizer therefore satisfies the equations in every coordinate. Conversely any fixed point has $B_j<1$, finite $p$, and these zero-gradient equations, so must equal that unique minimizer. Hence the Erlang fixed point approximation exists and is unique for fixed routing.

For integer requirements, the common generalized Erlang fixed point approximation usesMultiplication by $q_j$ gives the same equations for $\Phi$, so the existence and uniqueness argument also covers this generalized approximation. It remains an approximation, rather than the exact blocking law for a call requesting several units.

LetAt low density, $F(N)/N\to r$, so $r$ is the density-independent per-generation multiplication factor. The parameter $b$ has units of inverse population and sets the density scale at which crowding suppresses recruitment.

The fixed points satisfy $N=F(N)$. Besides extinction $N=0$, for $r>1$ there is the positive fixed pointSinceits multiplier isFor every $r>1$, $-1<F^{\prime }(N_{\ast })<1$, so fixed point stability for an iteration proves that $N_{\ast }$ is locally asymptotically stable.

Scale the population by $x_{\tau }=b{N}_{\tau }$. ThenThe map increases on $0<x<1$, decreases on $x>1$, and has global maximum $f(1)=r/4$. Starting from $x_1=1$ therefore givesFor $r>4$, writing $s=\sqrt{r}>2$ givesThus $1<L<N_{\ast }b=s-1<U$. Since $f$ decreases on $[L,U]$, $f(U)=L$ and $f(L)>f(s-1)=s-1>L$, while the global maximum gives $f(L)\le U$. Hence $f([L,U])\subseteq [L,U]$. The cobweb plot consequently remains in this invariant interval after the first iterate, and both endpoints are attained at $\tau =2,3$:The printed bound cannot include $\tau =1$, because $N_1=1/b<L/b$ when $r>4$.

Suppose that the continuous function $f:I\to I$ has no fixed point. Then $g(x)=f(x)-x$ is continuous and never zero, sois a continuous function from $I$ to $\{-1,1\}$. Since $f(-1)\ge -1$ and equality would be a fixed point, $h(-1)=1$. Similarly $h(1)=-1$. Multiplying $h$ by $-1$ contradicts part (c). Thus $f$ has a fixed point.

A fixed point satisfiesThe branch with $y=0$ isEvery other fixed point can be parametrized by its $x$-coordinate:Equivalently,The $x_{-}$ pair exists for $0\le a\le 4$, and the $x_{+}$ pair for $3\le a\le 4$.

The Jacobian matrix isOn $P_0(a)$ its eigenvalues are $1$ and $a(3-a)$, so this branch has a nonhyperbolic equilibrium at $a=0$ and $a=3$. On a nonzero-$y$ branch,which vanishes at $x=2$, hence at $a=4$ and $y=\pm \sqrt{2}$. The three bifurcation values and locations are therefore

For the center manifold calculation, append $\dot{a}=0$ as in the extended centre manifold for a parameter.

At $(a,x,y)=(0,0,0)$ put $\mu =a$ and write the centre manifold as $x=h(y,\mu )$. Its centre-manifold invariance equation isIt gives $h=\mu -y^2+O(3)$, and henceThis is the pitchfork bifurcation normal form with stable nonzero centre branches for $\mu >0$: the bifurcation at $a=0$ is supercritical.

At $(a,x,y)=(3,3,0)$ put $u=x-3$ and $\mu =a-3$. ThenAgain $u=h(y,\mu )=\mu -y^2+O(3)$, soThe nonzero centre branches exist for $\mu >0$ and are unstable in the centre direction, whereas $y=0$ is centre-stable there. Thus the bifurcation at $a=3$ is a subcritical pitchfork bifurcation with reversed normal-form parameter.

Finally fix $y_0=\pm \sqrt{2}$ and putThe translated system isThe extended centre manifold hasand thereforeThis is the saddle-node bifurcation normal form. For $\mu <0$ there are two nearby fixed points, which coalesce and disappear at $\mu =0$. The complete reductions are recorded in the Extended centre-manifold reductions of the 2019 Cambridge reflection-symmetric system.

If $x$ is a fixed point of $f$, thenConversely, applying the same argument to $f={\sigma }^{-1}g\sigma$ transports every fixed point of $g$ back to one of $f$. Thus the bijection $\sigma$ restricts to a bijection between the two fixed-point sets, so conjugate functions have the same number of fixed points.

A function $f:{\mathbb{R}}^m\to {\mathbb{R}}^r$ is differentiable at $x$ if there is a linear map $L:{\mathbb{R}}^m\to {\mathbb{R}}^r$ such thatThe map $L$, which is unique, is the derivative $f^{\prime }(x)$.

Expanding in a matrix increment $H$ givesTherefore $p$ is differentiable everywhere and

The inverse function theorem states that if $f$ is continuously differentiable near $a$ and $f^{\prime }(a)$ is invertible, then $f$ restricts to a bijection between neighbourhoods of $a$ and $f(a)$; its local inverse is continuously differentiable, with derivativeIn the normalized case $f(0)=0$ and $f^{\prime }(0)=I$, continuity of $f^{\prime }$ gives a closed ball ${\overline{B}}_r(0)$ on whichFor small $y$, define $T_y(x)=x-f(x)+y$. The mean value inequality makes $T_y$ a contraction mapping with constant $q$. If $\Vert y\Vert \le (1-q)r$, thenso $T_y$ maps the ball into itself. The contraction mapping theorem gives a unique fixed point $x_y$, and the fixed-point equation is precisely $f(x_y)=y$. Moreover,so $\Vert x_y-x_z\Vert \le \Vert y-z\Vert /(1-q)$. Thus $y\mapsto x_y$ is a continuous local inverse.

For the polynomial map in the question,The derivative is invertible, so the inverse function theorem supplies some $ϵ>0$ and a continuously differentiable $q:D_{ϵ}(I)\to {\mathcal{M}}_n$ such thatThere can be no inverse on all of ${\mathcal{M}}_n$, becauseso $p$ is not injective.

At a nonzero fixed point, both bracketed per-capita growth rates vanish:Hence the coexistence equilibrium of this Lotka-Volterra predator-prey model is

For $0<\mu <1$, one iterate sends the whole interval into $[1-\mu ,1]$, whereThis is a contraction mapping with unique fixed pointEvery orbit converges to $x_{\ast }$, alternating around it because the multiplier is $-\mu$.

For $\mu =1$, the first iterate lies in $[0,1]$ and $F(x)=1-x$ there. Hence $F^2$ is the identity on $[0,1]$: $1/2$ is fixed, while every other point in that interval belongs to the period-two orbit $\{x,1-x\}$.

Assume for contradiction that the involution $\varphi$ is fixed-point-free. The finite group action is then a covering space action, so the quotient mapis a double covering and $Y$ is an $n$-manifold.

Apply the long exact sequence in homology to the transfer chain map of a double covering. Since ${\mathbb{R}}^n$ is contractible, its positive-dimensional mod-two homology vanishes. The degree-zero portion, together with the fact that $q_{\ast }:H_0({\mathbb{R}}^n;{\mathbb{F}}_2)\to H_0(Y;{\mathbb{F}}_2)$ is an isomorphism, givesIn every higher degree the same exact sequence givesThus $H_i(Y;{\mathbb{F}}_2)\cong {\mathbb{F}}_2$ for every $i\ge 0$. This contradicts homology above the dimension of a manifold, which gives $H_i(Y;{\mathbb{F}}_2)=0$ for $i>n$. Therefore $\varphi$ has a fixed point.

The fixed point branches and their stability are as follows.

which is a negative-definite matrix, so this branch is stable.

Thus the $x$-versus-$\mu$ bifurcation diagram has the stable $x=0$ branch up to $\mu =1$, a stable branch $x=\sqrt{\mu -1}$ joining the bifurcation points $(1,0)$ and $(2,1)$, and the branch $x=\sqrt{\mu /2}$ changing from unstable to stable at $\mu =2$. The origin branch also lies at $x=0$ but remains unstable throughout.

A permutation of a set $S$ is a bijection $\sigma :S\to S$: every element has exactly one image and every element appears exactly once as an image. For a finite set, a permutation can also be recorded as an ordering of all its elements, once a reference order has been chosen. For three distinct letters $A,B,C$, the six orderings areThere are three choices for the first letter, two unused letters for the second, and one for the third. Every path in the tree gives one complete ordering, and every ordering has one such path, so the rule of product gives $3\cdot 2\cdot 1=3!=6$, a factorial.

Ordering versus mapping. Label the reference positions $1,\dots ,n$. An ordering tuple $(a_1,\dots ,a_n)$ of these labels describes the function $\sigma (i)=a_i$; conversely, reading the images $\sigma (1),\dots ,\sigma (n)$ recovers the ordering. For example,means $1\mapsto 3$, $2\mapsto 1$, $3\mapsto 4$, $4\mapsto 2$. Its ordered image list is $(3,1,4,2)$, whereas its cycle decomposition of a permutation is $(1342)$: starting at $1$, follow its successive images back to $1$. Parentheses in the latter expression encode arrows around a permutation cycle, rather than an image list in reference order. ^[ref]

The two viewpoints answer different useful questions about the same object. An ordering tells us which object occupies each position; a bijection tells us where each label is sent and allows successive rearrangements through function composition. All permutations of $n$ labels form the symmetric group, whose subgroups are permutation groups. The counting viewpoint supplies $n!$ possibilities, a factorial; the mapping viewpoint reveals how rearrangements combine.

Composition convention. Apply the rightmost permutation first:For the displayed $\sigma =(1342)$ and the transposition $\tau =(12)$, direct evaluation givesFor instance, $(\sigma \tau )(1)=\sigma (2)=1$, but $(\tau \sigma )(1)=\tau (3)=3$. Thus function composition of permutations need not be commutative; the omitted labels in these permutation cycles are fixed points.

Picard iteration replaces a tentative trajectory by the accumulated integral of its proposed rate. For an initial value problem $y^{\prime }=f(t,y)$, $y(t_0)=y_0$, define the Picard map and its starting function byHere $y_0(t)$ is the constant starting trajectory; the same symbol without its argument denotes the initial state. A fixed point of $\mathcal{P}$ solves this integral equation, and the fundamental theorem of calculus gives the original ordinary differential equation.

For $y^{\prime }=y$, $y(0)=1$, successive integrals giveAt $t=1$, the first five values are $1,2,2.5,2.66666667,2.70833333$, while $e\approx 2.71828183$. The fourth polynomial misses by about $0.00994850$. On $0\le t\le 1$, every omitted term of the exponential series is nonnegative and increasing, so the largest error occurs at $t=1$. The tail tends to zero, giving uniform convergence to the exponential function on this interval.

The beauty of Picard iteration is that each proposed trajectory supplies a better rate history for the next one. For this example, one integration adds exactly the next term of the exponential series. In general, the integrals may be difficult to compute, but the construction still explains local existence and uniqueness.

The short contraction mapping argument needs two independent controls. Choose a time-state cylinder $[t_0-a,t_0+a]\times {\overline{B}}_r(y_0)\subset U$ on whichThe second bound is the uniform local Lipschitz condition. Set $J=[t_0-h,t_0+h]$ with $0<h\le a$, $Mh\le r$ and $Lh<1$. In the function space of continuous functions $J\to {\mathbb{R}}^d$, use the supremum norm $\Vert u{\Vert }_{\infty }={\sup }_{t\in J}\Vert u(t)\Vert$. On its closed ball $\Vert u-y_0{\Vert }_{\infty }\le r$, direct integral estimates giveThus the map preserves the closed ball and is a contraction mapping there. This closed ball is a complete metric space, so the contraction mapping theorem gives convergence to exactly one fixed point. Any other local solution of a differential equation stays in the same ball after shortening its interval, and hence agrees there.

For $y^{\prime }=y$, take $r=1$, so $|y|\le 2$ in the state ball and $M=2$, $L=1$. Choosing $h=1/2$ gives $Mh=1$ and $Lh=1/2$, with the actual solution satisfying $|e^{1/2}-1|\approx 0.64872127<1$. The plotted interval $[0,1]$ is longer than this simple guarantee: the explicitly computed exponential series converges there anyway. The condition $Lh<1$ is sufficient for this one-step contraction mapping argument, rather than a necessary lifespan bound.

This map equals an explicit unit-step update of a logistic predator-prey differential equation. Its coexistence fixed point is $(b,(1-a-b)/a)$, requiring $a+b<1$. Its trace and determinant are $T=2-b/a$ and $D=(1-2b)/a$. The Schur stability criterion gives stability precisely when $b>(1-a)/2$ and $b<a+1/3$. The coexistence flip at $b=a+1/3$ is subcritical on $1/9<a<1/3$, with effective cubic coefficient $-9/[(3a+1{)}^2(9a-1)]$ when the $-1$ eigenvector has first component one. The unit-circle crossing $a+2b=1$ is a supercritical Neimark–Sacker bifurcation away from the strong resonances of a planar map $a=1/7,1/5$; in the eigenvector normalization $q_1=1$ its cubic radial coefficient is $-1/[a^2(1-a)]$. The whole nonnegative quadrant is not forward invariant, because $x^{\prime }$ can become negative. Local fixed-point stability is therefore not a global population-dynamics classification.

// Destination: dynamical-systems.bigb

The fixed point relation is $\mu =z-A\mathrm{sgn}(z)|z{|}^{\delta }$. Setting the fixed-point multiplier $A\delta |z{|}^{\delta -1}$ equal to one gives the two saddle-node bifurcations at $z=\pm r$. For fixed $0<A<1$, as $\delta \uparrow 1$ from below their parameter width tends to zero as $(1-\delta )A^{1/(1-\delta )}/e$, forming an exponentially narrow gluing-map cusp. Three scalar fixed points coexist inside this region: two stable outer branches and one unstable inner branch. For $\delta >1$ near one the formal saddle-node bifurcations recede to infinity, outside the local gluing-map domain.

Every continuous self-map of a nonempty compact convex subset of a locally convex space has a fixed point. Compactness is in the locally convex topology, which may be weaker than a norm topology. This version permits fixed-point arguments on a weak-star topology compact set of linear functionals. See Hartman's fixed-point chapter, Theorem 0.2.

For $x^{\prime }=-\mu \mathrm{sgn}(x)+A\mathrm{sgn}(y)|x{|}^{\delta }$, $y^{\prime }=\mathrm{sgn}(x)$, the invariant sign sheet has $y=\pm 1$ after one iterate. There $z=-xy$ obeys the displayed one-dimensional map. A negative fixed $z$ lifts to two symmetry-related fixed points, while a positive fixed $z$ lifts to one period-two orbit. The origin of the return map represents the separatrix, where the flow return time diverges, and is not an ordinary finite-period orbit of a group action.

This criterion holds even for reducible nonnegative matrices. If a fixed point $p\ne 0$ exists, $1-e^{-z}<z$ for $z>0$ gives $(\Lambda p{)}_i>p_i$ on its support. Some $\alpha >1$ therefore satisfies $\Lambda p\ge \alpha p$, forcing the spectral radius above one by the Perron–Frobenius theorem. Conversely take $v\ge 0$, $v\ne 0$, with $\Lambda v=\rho v$ and $\rho >1$. For sufficiently small $\eta >0$, $1-e^{-\eta \rho v_i}\ge \eta v_i$. Thus $\eta v$ is a subsolution; monotone iteration under $T$ gives a positive fixed point below $1$. Strictness of $1-e^{-z}<z$ excludes survival at the critical value for this Poisson offspring law.